# How do you find the vertex and intercepts for y = 3(x - 2)^2 + 5?

Dec 30, 2017

Vertex: $\left(2 , 5\right)$
y-intercept: $\left(0 , 17\right)$
x-intercept is undefined in ℝ

#### Explanation:

First, expand $y = 3 {\left(x - 2\right)}^{2} + 5$ into $y = 3 {x}^{2} - 12 x + 17$

The x-vertex of a quadratic expression ($y = a {x}^{2} + b x + c$) is given by $- \frac{b}{2 a}$.

So, the x-vertex for $y = 3 {x}^{2} - 12 x + 17$ will be $\frac{- \left(- 12\right)}{2 \cdot 3} = \frac{12}{6} = 2$

Plugging in the x-vertex into our original expression gives us the y-vertex, which is $3 \cdot 4 - 24 + 17 = 5$.

So, our vertex is at $\left(2 , 5\right)$.

The y-intercept is given when $x = 0$ in the expression, so it is $17$.

y-intercept: $\left(0 , 17\right)$

graph{y=3(x-2)^2+5 [-11.26, 11.545, 0.32, 11.72]}

From this graph of $y = 3 {x}^{2} - 12 x + 17$, it never hits the x-axis, so there is no x-intercept.